如何在python中创建自定义数字列表

Question

I need to create txt files containing a custom list based on a specific pattren in python or any way faster any program or program language with for loop or any loop i cant doing that because i want to create 639936 files example to explaining txt files containing this

1000001_0001
1000001_0002
1000001_0003
1000001_0004
1000001_0005
1000001_0006
1000001_0007
1000001_0008
--
--
--
1000001_9996
1000001_9997
1000001_9998
1000001_9999

---

1000002_0001
1000002_0002
1000002_0003
1000002_0004
--
--
--
1000002_9999

another files change just loop from 1000001 to 9999999

9999999_0001
9999999_0002
9999999_0003
9999999_0004
--
--
--
9999999_9999

the first number start 1000001 end 9999999 the another number start 0001 end 9999 between that underscore _ list this 1000001_0001 i want to spilit files to create multiple files becuase too large

for i in range(1000001,9999999):
for y in range(0,9999):
    cv = "{:07d}".format(i)
    vc = format(y)
    print (cv+"_"+vc)
    
    with open('1000001.txt','ab') as f:
        f.write(cv+"_"+vc+"\n")

Answer 1

I am assuming that you are trying to write multiple files, each file containing a set of numbers with 2 parts separated by an underscore, and in each file you have a fixed first number and changing (incrementing) "second number" from 0000 to 9999, and different files have different "first numbers" from 1000001 to 9999999.

If you are doing this in Python, List Comprehension is your friend. With list comprehension, you can create a list of items with any pattern, with a single command.

custom_list = ['1000001_' + str('%04d') % x for x in range(10000)]

# output: list of numbers as strings

Code explanation:- Variable x is looped from 0 to 9999 (10000-1). In each iteration, the x value is passed into function str() to convert integer into string format (you can't have underscore in a number, hence converting to string). '%04d' makes sure each x value is of 4 digits, by adding zeroes in front when needed. Each such strings made ('0000', '0001', ...) is then concatenated to the first number as string.

You can run another loop to loop through the first numbers and put this above written loop inside that, and make the outer loop save each list into a new file in each iteration (like converting into numpy array and using numpy.save). If you don't want to save these numbers into different files, you can even convert the above function into a nested list comprehension and make a 2D list.

Cheerio!

Edit: (adding code after OP's comment)

If you don't have numpy installed before you run this code, make sure you do.

# importing functions from numpy
from numpy import save

number_list = []
number_of_files = 1
for (index, i) in enumerate(range(1000001, 10000000)):
  small_list = [str(i) + '_' + str('%04d') % x for x in range(10000)]
  number_list += small_list

  # "save-test":
  # checking if number_list has enough elements for 1 file,
  # then saves it and empties the variable
  if (index + 1) % 9001 == 0:
    save('number_list' + str((index + 1) // 9001), number_list)
    number_list = []
    number_of_files += 1

# adding last file contents manually if it doesn't clear above "save-test"
if len(number_list) > 0:
  save('number_list' + str(number_of_files), number_list)

This will generate 1000 files, with 9001 * 10000 values in all files except last, which contains lesser number of values.

I don't know what you are trying to do with this code, but this should give you the results you mentioned in the comments.